barcode generator for ssrs If this constraint is satisfied, the output of the down-sampler has a DTFT as shown below. in Software

Drawing Code 128 Code Set A in Software If this constraint is satisfied, the output of the down-sampler has a DTFT as shown below.

If this constraint is satisfied, the output of the down-sampler has a DTFT as shown below.
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Going through the D/C converter produces the signal y o ( [ ) ,which has the Fourier transform shown below.
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Therefore, in order for y o ( / ) to be equal to x u ( / ) ,we require that
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MTl 5 1 /2000 in order to avoid aliasing. Tz = M T I to prevent frequency scaling.
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(6) With TI = Tz = 1 /20000 and M = 4, note that
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Therefore, there is no aliasing. Thus, as we see from the figure above,
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Digital audio tape (DAT) drives have a sampling frequency of 48 kHz, whereas a compact disk (CD) player operates at a rate of 44.1 kHz. In order to record directly from a CD onto a DAT, it is necessary to convert the sampling rate from 44.1 to 48 kHz. Therefore, consider the following system for performing this sample rate conversion:
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CHAP. 31
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Find the smallest possible values for L and M and find the appropriate filter H(eJu)to perform this conversion.
Given that 48000 = 2' . 3 . 5' and 44100 = 22 . 32 . S2 . 72, to change the sampling rate we require
Therefore, if we up-sample by L = 160 and then down-sample by M = 147, we achieve the desired sample rate conversion. The low-pass filter that we require is one that has a cutoff frequency
wc = min(Tr
= -160
and the gain of the filter should be equal to L = 160.
Suppose that we would like to slow a segment of speech to one-half its normal speed. The speech signal s a ( t ) is assumed to have no energy outside of 5 kHz, and is sampled at a rate of 10 kHz, yielding the sequence
s ( n ) = s,(nT,)
The following system is proposed to create the slowed-down speech signal.
Assume that S a ( j Q ) is as shown in the following figure:
(a) Find the spectrum of v ( n ) . (b) Suppose that the discrete-time filter is described by the difference equation
Find the frequency response of the filter and describe its effecl on v ( n ) .
(c) What is Y a ( j Q ) in terms of X , ( j Q ) Does y o ( ! ) correspond to slowed-down speech
(a) Since so(,) is sampled at the Nyquist rate, the DTFT of the sampled speech signal, s(n), is as follows:
WeJw)
Up-sampling by a factor of 2 scales the frequency axis of S(el") by a factor of two as shown below.
SAMPLING
[CHAP. 3
The unit sample response of the discrete-time filter is
which has a frequency response
H ( e J W= I )
+ cos w
To see the effect of this filter on v ( n ) ,note that due to the up-sampling, v ( n ) = 0 for n odd. Therefore. with
it follows that
y(n) =
u(n iv(n-I)+;u(n+l)
n odd
neven
Thus, the even-index values of v ( n ) are unchanged, and the odd-index values are the average of the two neighboring values. As a result, h ( n ) performs a linear interpolation between the values of v(n). The output of the DC converter, y,,(t), has a Fourier transform
Y,(.W) =
T T Y ( e J R T r ) IS21 < TIT,
otherwise
Since and
Y (ej") = H (pi''')v (ei") = ( I
+ cos w
)(ej") ~
v ( e J w ) s(eJ2") =
then which is the product of (1
+ cos QT,) and T , S ( ~ J ' " ~as%illustrated below. )
Thus, y,(t) does not correspond to slowed-down speech due to the images of s,,(r) that occur in the frequency range 5000n < IS21 < IOOOOn and the nonideal linear interpolator. Note that a better approximation would be to use a DC converter with a sampling rate of 2T, to eliminate the images.
Shown in the figure below are two different ways of cascading an up-sampler with a down-sampler.
CHAP. 31
SAMPLING
(a) If M = L , show that the two systems are not identical. (b) Under what conditions will the two systems be identical
(a) In the first system, which consists of an up-sampler followed by a down-sampler, note that w l ( n )is a sequence
that is formed by inserting L - I zeros between each value of x ( n ) . The down-sampler then extracts every Lth value of w l ( n ) ,thus producing the output
In the second system, however, the down-sampler extracts every Lth sample of s ( n ) and discards the rest. The up-sampler then inserts L - I zeros between each value of w 2 ( n j . Thus,
Therefore, the two systems are not the same.
(b) In order to analyze these systems when L following figure:
# M, note that y z ( n ) in the second system has the form shown in the
On the olher hand, the sequence w , ( n ) in the first system is as shown below.
Note that y l ( n ) is formed by extracting every Mth value of w l ( n ) ,
SAMPLING Clearly, so yl(kL) = wl(kML) = x(kM) yl(kL) = yz(kL)
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